Numerical Ability Test - 30

Given:

The probability that A hits a target is 1/3 and the probability that B hits the target is 2/5

Calculation:

Let E₁ be the event that A hits the target

and E₂ = event that B hits the target

Then, P(E₁) = 1/3 and P(E₂) = 2/5

Clearly, E₁ and E₂ are independent events.

Now,

P(E₁∩E₂) = P(E₁) × P(E₂)

⇒ (1/3 × 2/5)

⇒ 2/15

Now,

P(target is hit) = P(A hits or B hits)

⇒ P(E₁ ∪ E₂) 

⇒ P(E₁) + P(E₂) − P(E₁ ∩ E₂)

⇒ (1/3 + 2/5 − 2/15)

⇒ 9/15

⇒ 3/5

∴ The required probability is 3/5.

Concept used:

By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that:  P(E∪F) = P(E) + P(F) – P(E∩F)

Calculation:

In a deck of 52 cards there are 2 colours. Each colour having 26 cards.

As we need to choose 4 cards out of 52. Let S represents the sample space.

So, n(S) = ⁵²C₄ 

Let A represents the event that all 4 cards drawn are black

So, n(A) = ways in which 4 cards can be selected from 26 black cards. 

⇒ n(A) = ²⁶C₄

Let B represents the event that all 4 cards drawn are red

So, n(B) = ways in which 4 cards can be selected from 26 Red cards. 

⇒ n(B) = ²⁶C₄

As we need to find the probability of the event such that all drawn cards are from the same color. This means we need to find  P(A∪B)

P(A∪B) = P(A) + P(B) – P(A ∩ B)

As both events, A and B have no common elements or we can say that they are mutually exclusive

So, P(A ∩ B) = 0

So, P(A ∪ B) = P(A) + P(B)

Concept:

Combination formula:

The formula of a combination of r objects out of n objects is given as follows:

Permutation formula:

The formula for permutation of r objects out of n objects is given as follows:

Given:

In a class, there are 15 boys and 10 girls.

Calculation:

Number of boys = 15

Number of girls = 10

Total number of ways of selecting 3 students from 25 students = ²⁵C₃

Number of ways of selecting 1 girl and 2 boys = selecting 2 boys from 15 boys and 1 girl from 10 girls Number of ways in which this can be done = ¹⁵C₂ × ¹⁰C₁

So, required probability =  (¹⁵C₂ × ¹⁰C₁)/²⁵C₃

Concept used:

P(A) = n(A)/n(S)

P(A) is the probability of an event “A”

n(A) is the number of favourable outcomes

n(S) is the total number of events in the sample space

Calculation:

Number of days in leap year = 366

A week has 7 days and total days are 366.

⇒ Number of Sundays in a leap year = 366/7 = 52 Sundays + 2 days

So, Total outcomes with 2 days = (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday) = 7

Number of outcomes without Sundays = 5

⇒ Probability of leap year with 53 Sundays = 2/7

∴ Probability of leap year with 53 Sundays is 2/7.

Given:

Total number of tickets = 50

Number of tickets drawn = 2

Concept used:

Let A and B are independent events,

then P(A ∩ B)= P(A). P(B)

Calculation:

Prime number between 1 to 50 = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29,31, 37, 41, 43, 47} = 15

A : The event of getting a prime number in the first draw

B : The event of getting a prime number in the second draw

The event A and B are independent events.

P(A) = Probability of getting a prime number in the first draw = 15/50

⇒ 3/10

Number of remaining cards in pack = 49

P(B) = Probability of getting a prime number in the second draw = 14/49

⇒ 2/7 

The probability that both the tickets drawn have prime numbers P(A ∩ B) = P(A). P(B)

⇒ 3/10 × 2/7

Given:

A bag contains 6 white and 4 red balls

Concept used:

Probability of occurrence of the event, P(E) = (Number of favorable outcome, n(E))/(Number of possible outcome, n(S))

Calculation:

Here, total number of balls in the bag = 6 + 4 = 10

Number of ways in which 2 white balls can be drawn from 6 white balls and 1 red ball can be drawn from 4 red balls = n(E) = ⁶C₂ × ⁴C₁

⇒ 60

Total number of ways in which 3 balls can be drawn out of all the all the balls = n(S) = ¹⁰C₃ 

⇒ 120

So, Probability P(E) = n(E)/n(S)

⇒ 60/120 

⇒ 1/2

∴ The required answer is 1/2.

Formula used:

P(E) = n(E)/n(S)

Where, n(E) = Number of favourable outcome

n(S) = Total number of outcome

Calculation:

We know that in a simultaneous throw of two dice,

n(S) = 6 × 6

⇒ 36

Let E be the event of getting a total of 7

{(1, 6), (2, 5), (3, 4), (5, 2), (6, 1), (4,3)}

P(E) = n(E)/n(S)

⇒ 6/36

⇒ 1/6

∴ The probability of getting a total of 7 is 1/6.

Given:

2, 3, 3, 4, 4, 4, 5, 5, 5, 5

Concept used: 

Average = Sum of observation / Number of observations

Probability = Number of favorable outcomes / Number of possible outcomes

Calculation:

⇒ Number of observations = 10

⇒ Sum of observations = 2 + 3 + 3 + 4 + 4 + 4 + 5 + 5 + 5 + 5

⇒ 40

⇒ Number of favorable outcomes = 3

⇒ Number of possible outcomes = 10

Concept:

Binomial Distribution: Binomial distribution is a probability distribution used in statistics that states the likelihood that a value will take one of two independent values under a given set of parameters or assumptions.

The binomial distribution is the discrete probability distribution that provides only two possible results in analysis, i.e., either success or failure.

Binomial Distribution Formula:- 

P(x) = ⁿC_x p^x q^{n - x}

where, n = number of trials 

x = the number of success desired

p =probability of getting a success in one trial

q = 1 - p = probability of getting a failure in one trial

Calculation:

Eight coins are tossed together.

⇒ n = 8

∴ If eight coins are tossed together, then the probability of getting exactly 3 heads is 7/32.

Given:

If Mohan speaks the truth 70% of the times, Sohan speaks the truth 50% of the times.

Calculation:

Probability of Mohan speaking truthP(Mohan) = 70% = 70/100 = 7/10

Probability of Sohan speaking truthP(Sohan) = 50% = 50/100 = 1/2

Probability of two mutually exclusive events to occur simultaneously = P(Mohan) × P(Sohan)

⇒ 7/10 × 1/2

⇒ 7/20

⇒ 0.35

∴ Required probability is 0.35